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Linear algebra proof question

  1. Dec 14, 2008 #1
    the question is in this link?


    i dont know how to prove it
    i know basic stuff
    that Ker T is all the vectors which go to line of zeros in matrix row reduction.
    Im T is produced when i traverse the matrix and do a ow reduction
    and the remaining(non zero) vectors are Im T
    i also know that V=Im T + Ker T

  2. jcsd
  3. Dec 14, 2008 #2


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    The question appears to be asking "Is there a linear transformation T, from R3 to R3 such that ker T is a subset of the inverse of T and such that the kernel of T is spanned by (1, 1, 0). I may be misunderstanding since what I am interpreting as "inverse of T" is shown as "T-1" and if that is so then the problem is trivial: in order to have an inverse, T must be one-to-one- in other words ker T must be {(0,0,0)}. The second questions ask the same about a linear operator S but has kernel of T spanned by (3, 1, 0) and (2, 0, 0). That seems too trivial a question. I may be misunderstanding "T-1" and "S-1".
  4. Dec 14, 2008 #3
    sorryy disregard -1 thing
    in my language its the word for "AND"
    so its first expression and second expression
  5. Dec 14, 2008 #4


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    Well, any linear function from R3 to R3 maps (x, y, z) into (ax+ by+ cz, dx+ ey+ fz, gx+ hy+ iz) so finding a linear function is just a matter of finding 9 numbers. You are told that the kernel is spanned by (3, 1, 0) and (2, 0, 0) so you know that 3a+ b= 0, 3d+ f= 0, 3g+ h= 0, 2a= 0, 2d= 0, and 2g= 0. That should help a lot! Since the kernel is 2 dimensional, you know that the image must be only 1 dimensional: all columns are multiplies of each other. Finally, you want that one dimensional image to be a subspace of the kernel.
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